Caculate $\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$, with D:$\frac{x^2}{2}+y^2\leq1$ I found some difficulty with this exercise:
Calculate
$$\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$
with $D := \left\{(x,y)\in\mathbb{R}^{2}\mid \dfrac{x^2}{2}+y^2\leq1\right\}$
I use change of Variables in Polar Coordinates, but the integral become so hard to calculate.
I think maybe we change variables $u = x^2 + y^2$, the integral will be easier, but I can't find $v(x,y$) to have the Jacobi easy to calculate.
 A: Take $$I=\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$
Consider the change of variables $(x,y)\to(r\cos\theta,r\sin\theta)$.
Let $\partial D=\{(x,y)\in\mathbb{R}^2:\frac{x^2}2+y^2=1\}$. Then for all $(x,y)=(r\cos\theta,r\sin\theta)\in\partial D$, $r^2(1-\frac12\cos^2\theta)=1$. Hence,
$$I=\int_0^{2\pi}\left(\int_0^{R(\theta)}\frac{r^2}{\sqrt{4-r^4}}rdr\right)d\theta$$
where $R(\theta)=\frac1{\sqrt{1-\frac12\cos^2\theta}}$. Now,
$$I=\int_0^{2\pi}\frac12\left(2-\sqrt{4-[R(\theta)]^2}\right)d\theta$$
$$\implies I=2\pi-\frac12\int_0^{2\pi}\sqrt{4-\left(1-\frac12\cos^2\theta\right)^{-2}}d\theta$$
A: Here is a possible way to evaluate explicitly.
We use the change of variables $x=r\sqrt 2\; \cos t$, $y=r\sin t$. Then we have formally
$$
\begin{aligned}
dx &= \sqrt 2\; \cos t\; dr - r\sqrt 2\; \sin t\; dt\ ,\\
dy &= \sin t\; dr + r\; \cos t\; dt\ ,\\
dx\wedge dy &= 
dr\wedge dt\cdot
\sqrt 2\; \cos t\cdot r\; \cos t
- dt\wedge dr\cdot r\sqrt 2\; \sin t
\cdot
\sin t
\\
&=
dr\wedge dt\cdot r\sqrt 2\cdot (
\cos^2 t +\sin^2 t)
\\
&=
dr\wedge dt\cdot r\sqrt 2\ .\\[2mm]
x^2 + y^2
&=2r^2\cos ^2t + r^2\sin^2 t\\
&=r^2(1+\cos^2 t)\ .
\end{aligned}
$$
So the given integral $I$ can be computed as follows:
$$
\begin{aligned}
I 
&=\iint_D \frac{x^2+y^2}{\sqrt {4-(x^2+y^2)^2}}\; dx\; dy
\\
&=4
\iint_{substack{(x,y)\in D\\x,y\ge 0}}
\frac{x^2+y^2}{\sqrt {4-(x^2+y^2)^2}}\; dx\; dy
\\
&=
4
\iint_{\substack{0\le r\le 1\\0\le t\le \pi/2}}
\frac{r^2(1+\cos^2 t)}{\sqrt{4-r^4(1+\cos^2 t)^2}}
\;r\sqrt 2\; dr\; dt
\\
&\qquad\text{ ... now use $s=r^4$, $ds=4r^3\; dr$}
\\
&=
\sqrt 2
\iint_{\substack{0\le s\le 1\\0\le t\le \pi/2}}
\frac
{\color{blue}{1+\cos^2 t}}
{\sqrt{4-s\color{red}{(1+\cos^2 t)^2}}}
\;ds\; dt
\\
&\qquad\text{ ... now use 
$\int_0^1\frac{ds}{\sqrt{4-s\color{red}{a}}}
=\frac 1{\color{red}{a}}(4-2\sqrt {4-\color{red}{a}})$}
\\
&=
\sqrt 2
\int_0^{\pi/2}
\frac{\color{blue}{1+\cos^2 t}}{\color{red}{(1+\cos^2 t)^2}}
\left(4-2\sqrt{4-\color{red}{(1+\cos^2 t)^2}}\right)
\; dt
\\
&=
2\sqrt 2
\int_0^{\pi/2}
\frac1{1+\cos^2 t}
\left(2-\sqrt{4-(1+\cos^2 t)^2}\right)
\; dt
\\
&=
2\pi
-
2\sqrt 2
\int_0^{\pi/2}
\frac1{1+\cos^2 t}
\sqrt{2^2-(1+\cos^2 t)^2}
\; dt
\\
&=
2\pi
-
2\sqrt 2
\int_0^{\pi/2}
\frac1{1+\cos^2 t}
\sqrt{(3+\cos^2 t)(1-\cos^2 t)}
\; dt
\\
&=
2\pi
-
2\sqrt 2
\int_0^{\pi/2}
\frac1{1+\cos^2 t}\cdot
\sqrt{3+\cos^2 t}
\; \sin t\; dt
\\
&\qquad\text{ ... now use $u=\cos t$}
\\
&=
2\pi
-
2\sqrt 2
\int_0^1
\frac1{1+u^2}\cdot
\sqrt{3+u^2}
\; du
\\
&=
\color{forestgreen}{
2\pi
-
\sqrt 2\log 3
-
4\arctan\frac 1{\sqrt 2}}\ .
\end{aligned}
$$
$\square$

Numerical check:
sage: 4 * numerical_integral( lambda x:
....:        numerical_integral( lambda y :
....:            (x^2 + y^2) / sqrt(4 - (x^2 + y^2)^2),
....:                            (0, sqrt(1-x^2/2)) )[0],
....:                         (0, sqrt(2)) )[0]
2.2675940740738505
sage: ( 2*pi - sqrt(2)*log(3) - 4*atan(1/sqrt(2)) ).n()
2.26759407407385

